Homework Help:
Force on a water container

Ok there is a dam of width "b" and water is fille don one side of it to a height h . Now i need to calculate the force on the wall of the dam on the side where water is filled .
This is what I did:

I considered a small layer of water of dx (vertical) at a distance x from top of water surface .

Then:

[itex] \int dF= \int \delta xg (bdx)[/itex]

Limits 0--->h on RHS

[itex] F=\frac {b \delta gh^2}{2} [/itex]

Therefore the force on the wall depends on :

Width of the dam
The height "h" to which water is filled
Density of water
and g

But one thing i noticed is that the force on the wall doesnt depend on the length of the river..!

2) Now suppose i want to calculate force due to water ina cuboid lying horizontally on floor ..Its length is l breadth is b and height h ... now if it lies on floor with breadth as its height...then it means as per previous conclusion that length of the cuboid lying horizontally wouldnt make any difference...So this means i keep on increasing its length without any effect on force due to water on the front face?

Ok there is a dam of width "b" and water is fille don one side of it to a height h . Now i need to calculate the force on the wall of the dam on the side where water is filled .
This is what I did:

I considered a small layer of water of dx (vertical) at a distance x from top of water surface .

Then:

[itex] \int dF= \int \delta xg (bdx)[/itex]

Limits 0--->h on RHS

[itex] F=\frac {b \delta gh^2}{2} [/itex]

Therefore the force on the wall depends on :

Width of the dam
The height "h" to which water is filled
Density of water
and g

But one thing i noticed is that the force on the wall doesnt depend on the length of the river..!

2) Now suppose i want to calculate force due to water ina cuboid lying horizontally on floor ..Its length is l breadth is b and height h ... now if it lies on floor with breadth as its height...then it means as per previous conclusion that length of the cuboid lying horizontally wouldnt make any difference...So this means i keep on increasing its length without any effect on force due to water on the front face?

Yes, that's exactly what it means. The pressure water exerts on a surface depends only on the height of the water above that surface.

[itex] F=\frac {b \delta gh^2}{2} [/itex]
But one thing i noticed is that the force on the wall doesnt depend on the length of the river..!

I don't know if I'm right here, but in your calculations you accounted for the force caused by the hydrostatic pressure of the water held by the dam. If it is a still body of water, then this is the only pressure on the dam and that's it, it doesn't depend on the size of the body of water, just on the depth next to the dam.

But if you're holding off a river that's slamming into the dam, then you have to deal not only with hydrostatic pressure, but also with kinetic energy of the river. Which probably will depend on speed of the river, it's length, angle of fall, etc.

Once more, this is not my field of science, just my reasoning about the problem

Staff: Mentor

But one thing i noticed is that the force on the wall doesnt depend on the length of the river..!

That's right. The water pressure depends only on the depth of the water, not on how much water there is.

2) Now suppose i want to calculate force due to water ina cuboid lying horizontally on floor ..Its length is l breadth is b and height h ... now if it lies on floor with breadth as its height...then it means as per previous conclusion that length of the cuboid lying horizontally wouldnt make any difference...So this means i keep on increasing its length without any effect on force due to water on the front face?

If I understand you correctly, yes.

Let me rephrase it. Let's say you have a cuboid with sides a (x-direction), b (y-direction), c (z-direction = vertical). The net force on vertical side b-c depends on b and c, but not on a. (I think that's what you are saying.) The net force due to water pressure on any vertical surface does not depend on the dimension of the cuboid perpendicular to that surface. (As long as changing that dimension has no effect on the water pressure.)

Ok, one thing more... I think the length of the cuboid has an indirect on the force on the front wall..because if we keep the volume fixed, then if we increase the length , then the height of the water decreases and the force decreases ...therefore keeping the volume fixed , the force on wall is inversely proportional to the length of the cuboid.

Staff: Mentor

Dr.Brain said:

Ok, one thing more... I think the length of the cuboid has an indirect on the force on the front wall..because if we keep the volume fixed, then if we increase the length , then the height of the water decreases and the force decreases ...therefore keeping the volume fixed , the force on wall is inversely proportional to the length of the cuboid.

Not exactly. It depends on how much the water level increases. Which is why I added the statement: "As long as changing that dimension has no effect on the water pressure."

In a large body of water, no problem: the increased height of the water could be neglected; in a small volume of water, not so.